MCQ
જો $f(t) = \int_{\, - t}^{\,t} {\frac{{dx}}{{1 + {x^2}}},} $ તો $f'(1) = . . . ..$
- A$Zero$
- B$2/3$
- C$ - \,1$
- ✓$1$
$ = [{\tan ^{ - 1}}x]_{ - t}^t$$ = 2{\tan ^{ - 1}}t$
Differentiating with respect to $t$,
$f'(t) = \frac{2}{{1 + {t^2}}}$
==> $f'(1) = \frac{2}{2} = 1$.
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